An object is thrown vertically from a height of #5 "m"# at #3 "m/s"#. How long will it take for the object to hit the ground?
1 Answer
Answer:
Explanation:
"Thrown vertically" could mean either upwards or downwards. I am assuming it to be downwards.
Step 1: Understand the problem.
An object is under free fall. It is accelerating at
Step 2: Identify the relevant equations.
To find displacement,
#s = v_0 t + 1/2 a t^2#
To get the time as a function of the distance traveled, make
#t^2 + frac{2 v_0}{a}t  frac{2 s}{a} = 0#
Subsequently, complete the square.
#(t + frac{v_0}{a})^2  frac{2 s}{a} = (frac{v_0}{a})^2#
#(t + frac{v_0}{a})^2 = frac{v_0^2}{a^2} + frac{2 s}{a}#
#= frac{v_0^2 + 2as}{a^2}#
Reject the negative square root and take the positive one. This is because we only restrict ourselves to
#t + frac{v_0}{a} = frac{sqrt{v_0^2 + 2as}}{a}#
#t = frac{sqrt{v_0^2 + 2as}  v_0}{a}#
Step 3: Set up the coordinate system.
I let the downwards direction be positive. The ground level is arbitrarily set to be the origin (i.e.
Step 4: Identiy the parameters.
In this question,

#a = g = 9.81 quad "m/s"^2# 
#v_0 = 3 quad "m/s"# 
#s = 5 quad "m"#
Step 5: Plug into the equation.
#t = frac{sqrt{v_0^2 + 2as}  v_0}{a}#
#= frac{sqrt{(3 quad "m/s")^2 + 2(9.81 quad "m/s"^2)(5 quad "m")}  (3 quad "m/s")}{9.81 quad "m/s"^2}#
#= 0.749 quad "s"#
Now earlier, I assume that the object has an initial velocity of
#v_0 =  3 quad "m/s"#
Plugging in the values again
#t = frac{sqrt{v_0^2 + 2as}  v_0}{a}#
#= frac{sqrt{(3 quad "m/s")^2 + 2(9.81 quad "m/s"^2)(5 quad "m")}  (3 quad "m/s")}{9.81 quad "m/s"^2}#
#= 1.361 quad "s"#